Question

Помогите решить 1,2,3,7,5 Срочно. Заранее спасибо.))

Answer 1

4х^2-12=0
4х^2=12 (:4)
х^2=3
х=под корнем 3

Answer 2

1.
4x²-12=0
4(x²-3)=0
x²-3=0
x²=3
x₁= -√3
x₂=√3
Ответ: -√3; √3.

2.
[4x/(x-y)(x+y)] - [4(x-y)/(x+y)(x-y)]=(4x-4x+4y)/(x²-y²)=(4y)/(x²-y²)

3.
3x>12+11x
3x-11x>12
-8x>12
x<12 : (-8)
x< -1.5

5x-1<0
5x<1
x<0.2

{x< -1.5
{x<0.2
\\\\\\\\\\\
-------- -1.5 ---------- 0.2 ------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x< -1.5
x∈(-∞; -1.5)

5.
3x²+2x-1=3x²+3x-x-1=3x(x+1)-(x+1)=(x+1)(3x-1)

7.
sin²α-2sinαcosα+cos²α+2sinαcosα=sin²α+cos²α=1